Perfect if and only if Triangular

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Advances in Theoretical and Applied Mathematics ISSN 0973-4554 Volume 1, Number 1 (017), pp. 39-50 Research India Publications http://www.ripublication.com Perfect if and only if Triangular Tilahun Muche, Mulatu Lemma, George Tessema and Agegnehu Atena Department of Mathematics, Savannah State University, Savannah, GA, 31404, U.S.A. Abstract A number n is perfect when σ(n) = d = n. It was Euclid who 0 d n d n proved that if ( k 1) is a prime number, Mersenne prime, then N = k 1 ( k 1) is an even perfect number. Moreover, if N is an even perfect number then N = T m for some m N and m 3 is a triangular number m where T m = i=1 i. In this paper we proved the necessary and sufficient condition for an even triangular number T m to be a perfect number N= k 1 ( k 1)besides T m 4 mod 10 and T m mod 10. Keywords: Perfect Numbers, Triangular Numbers and Mresenne Primes. Mathematical subject Classification: 11B7, MSC 010 INTRODUCTION A Perfect Number is a positive integer with the property that it coincides with the sum of all its positive divisors other than the number itself [1]. Thus, an integer n 1 is a perfet number if 0 d<n d n d = n

40 Tilahun Muche, Mulatu Lemma, George Tessema and Agegnehu Atena The nth triangular number is the number of dots composing a triangle with n dots on a side, and is equal to the sum of the n natural numbers from 1 to n. [] n i = i=1 n(n + 1) Example 1: Triangular Numbers: 1, 3, 6,10,15,1,8,36,45... Perfect Numbers: 6, 8,496, 816, 33550336, 8589869056,13743869138, 3058430081399518, 6584559915698317446546961595384176,... The number 6 is unique in that 6 = 1 + + 3 where 1, and 3 are all of the proper divisors of 6. The number 8 also shares this property, for 8 = 1+ +4 +7 + 14. These perfect numbers have been a great deal of mathematical study- indeed, many of the basic theorems of numbers theory stem from the investigation of the Greeks into the problem of perfect and Pythagorean numbers. The Pythagoreans introduced the name perfect and there are speculations that there could be religious or astrological origins because the earth was created in 6 and the moon needs 8 days to circle the earth, mystical associations are natural. The early Hebrews also studied perfect numbers [3]. Definition 1: The sum of divisors is the function σ(n) = the positive divisors of n including 1 and n itself. d n d, where d runs over Definition : The number n is called perfect if σ(n) = n, when σ(n) < n we say n is deficient, σ(n) > n we say n is abundant. Example : 6 and 8 are perfect as σ(6) = 1 + + 3 + 6 = 1 = (6) and σ(8) = 1 + + 4 + 7 + 14 + 8 = 56 = (8).

Perfect if and only if Triangular 41 Euclid was the first mathematician who categorized even perfect numbers. He noticed that 6= 1. 3 1 = 1 ( 1) 8 =. 7 = ( 3 1) 496 = 16. 31 = 4 ( 5 1) 816 = 64. 17 = 6 ( 7 1) Theorem 3 (Euclid)[4,9]: If ( n 1) is prime then N = ( n 1) is perfect. Proof: The only prime divisors of N are ( n 1) and. Since ( n 1) occurs as a single prime, we have that σ( n 1)= (1 + ( n 1))= n, and thus σ(n) = σ ( )σ ( n 1) = ( n 1 1 ) n = n ( n 1) =. ( n 1) = N So N is perfect. Mersenne primes: Monk Martin Mersenne, a colleague of Descartes, Fermat and Pascal created with investigating these unique primes as early as 1644. He knew ( n 1) is prime for n =, 3, 5, 7, 11, 13,17 and 19. [5, 6] Definition 4: A Mersenne prime is a prime number of the form M n = P n 1 where P n is a prime number. Proposition 5:[5] (Cateldi Fermat) If ( n 1) is prime, then n itself is prime. Proof: x n 1= (x 1) (x +... +x + 1). Suppose we can write n = rs where r, s > 1. Then n 1 = ( r ) s 1 = ( r 1)(( r ) s 1 +... + r + 1) so that ( r 1) ( n 1) which is prime, a contradiction. Theorem 6: If N is an even perfect number, then N = ( n 1) where( n 1) is prime. Proof: Since n m = ( n 1)σ(m), every prime divisor of ( n 1) must also divide m, for it is odd and cannot divide n. So, suppose p α divides ( n 1) with p prime.

4 Tilahun Muche, Mulatu Lemma, George Tessema and Agegnehu Atena From the fact that if a b, then σ(a) have, σ(m) a σ(b) b σ(pα ) 1+p+...+ p m pα = p 1 = σ(n) N = σ( )σ(m) n m where equality holds only if a = b we p 1 + p = 1+p p p (n 1)(1+p) n p. Hence = 1 + (n 1) p n p This is only satisfied when the fraction on the right is zero, so that p = ( n 1), = 1 and m = p. Hence N = ( n 1).. Proposition 7: Even perfect number ends in either 6 or 8. Proof: Every prime number p is of the form p = 4m + 3 or p = 4m + 1. In the former case, N = ( n 1) = 4m ( 4m+1 1) = (16) m (. (16) m 1) 6 m ((6) m 1) 6(mod10) Since by induction one can show that 6 m 6(mod10) for all m. Similarly in the latter case, N = ( n 1) = 4m+ ( 4m+3 1) = 4(16) m (8. (16) m 1) (4)(6)(8(6) 1) 4(8 1) 8(mod10). Finally, if n =, N = 1 ( 1) = 6 and so we have the result that even perfect number ends in either 6 or 8. MAIN RESULTS m Let A(x) = i=0 a i x i and B(x) = i=0 b i x i. Then A(x)B(x) = C(x) = n+m k=0 c k x k where k C k = i=0 a i b k i for 0 k m + n. n The decimal expansion of a positive integer N, 0 a k < 10 where a 0 is the unit digit of N is given by N = A(10) = a m 10 m + a m 1 10 m 1 +... + a 1 10 1 + a 0 10 0 = (a m a m 1 a m... a 1 a 0 ) 10 = m i=0 a i 10 i

Perfect if and only if Triangular 43 Theorem 8 [7]: A triangular number T m = i=1 i is even if and only if m = (4k 1) or m = 4k for some ε Z +. m Theorem 9: Even triangular numbers T m end not with or 4. That is neither T m 4 mod 10 nor T m mod 10. Proof: Suppose T m is an even triangular number. Then either m = (4k 1)or = 4k. a) Suppose m = 4k 1. This implies T m = T 4k 1 = (4k 1)( 4k) = 4k 1 i=1 i = k( 4k 1). LetA (10) = k = m i=0 a i 10 i = (a m a m 1 a m... a 1 a 0 ) 10 be decimal expansion of the factor(k) of an even triangular number T 4k 1 where the unit digit a 0 { 0,, 4, 6, 8}. Let b 0 be the unit digit of the factor B(10) = (4k 1) of T m where the decimal expansion is B(10) = 4k 1 = n i=0 b i 10 i = (b n b... b 1 b 0 ) 10. Consider T m = T 4k 1 = (k)(4k 1) = (k)( (k) 1) = A(10) B(10) = ( m i=0 a i 10 i ) ( n i=0 b i 10 i ) = C(10) n+m = i=0 c i 10 i = (c m+n c m+ c 1 c 0 ) 10 The constant term c 0 of C(10) = T m = T 4k 1 is c 0 = a 0 b 0. We consider each unit digit a 0 { 0,, 4, 6, 8} of k = A(10) to determine unit digits b 0 of B(10) = (4k 1) and c 0 of C(10). 1) a 0 = 0 c 0 = 0 ) a 0 =, b 0 = * 1 = 3 c 0 = 6 3) a 0 = 4, b 0 = *4 1= 7 a 0 b 0 = c 0 = 8 (Because (4)*(7) = 8 = 10 1 + 8 10 0 ) 4) a 0 = 6, b 0 = 1, because *6 1 = 11 = 1 10 1 + 1 10 0 c 0 = 6 5) a 0 = 8, the unit digit b 0 = 5 because *8 1 = 1 5 = 1 10 1 + 5 10 0 c 0 = 0.

44 Tilahun Muche, Mulatu Lemma, George Tessema and Agegnehu Atena b) Form = 4k, in similar approach one can show that an even triangular number T m has the following sequence of unit digits. Hence if a triangular number is even, then its unit digit is either 0, 6 or 8 but not and 4. This implies T m 4 mod 10 and T m mod 10. Proposition 10 [8]: Every even perfect number ends in either 6 or 8. Theorem 11: An even triangular number T 4k for each k 1cannot be written in the form of ( n 1) for any n.

Perfect if and only if Triangular 45 Proof: Suppose T 4k = and k 1. Then, 4k( 4k+1) = k( 4k + 1) = ( n 1) for some n k( 4k + 1) = ( n 1) iff 4k + k = n ( n 1) 4k + k - n ( n 1) = 0 (4k ( n 1))(k + n )= 0 4k = n 1 or k + n = 0 k = n 1 (not an integer) or k = n (not a positive integer) k ε Hence T 4k ( n 1) for any k 1 and n. Corollary 1: If an even triangular number T m is perfect, then m = (4k 1) for some k 1. 6 10 8 36 66 78 10 136 190 10 76 300 378 406 *3 *5 4*7 4*9 6*11 6*13 8*15 8*17 10*19 10*1 1*3 1*5 14*7 14*9 t 3 t 4 t 7 t 8 t 11 t 1 t 15 t 16 t 19 t 0 t 3 t 4 t 7 t 8 1 ( 1) 3 1 ( 3 1) 4 1 ( 4 1) 496 58 630 666 780 80 946 990 118 1176 136 1378 1540 1540 16*31 16*33 18*35 18*37 0*39 0*41 *43 *45 4*47 4*49 6*51 6*53 8*55 8*57 t 31 t 3 t 33 t 36 t 35 t 40 t 37 t 44 t 39 t 48 t 41 t 5 t 43 t 56 5 1 *( 5 1) Table III: Even Triangular Numbers with some in ( n 1) form.

46 Tilahun Muche, Mulatu Lemma, George Tessema and Agegnehu Atena Theorem 13: An even triangular number T m is perfect if and only if m = ( t 1) for some prime number t. Proof: A triangular number T m is even if and only if either m = (4k 1) or m = 4k for some k 1 and an every prime number p > is of the form p = (4l + 3) or p = 4l + 1. [7] Suppose T m is perfect. This implies T m = ( n 1) and ( n 1) is prime. But ( n 1) prime only if n is prime number. By (Theorem 11) and the later remark above the only choice for m is m = (4k 1) = 4(k 1) + 3 = 4l + 3 but not m = 4k. m 4k 1 Hence, T m = i=1 i = i=1 i = (k)(4k 1) = ( n 1) and T m = ( n 1) ( 4k 1)(k) = ( n 1) (1) It is easy to show that gcd(k, 4k 1) = gcd (,, n 1) = 1 We use () to show that (4k 1 ) = ( n 1). = gcd(k, n 1) = gcd(, 4k 1). () Because ( n 1)is prime, either gcd (4k 1, n 1) = 1 or ( n 1) (4k 1 ). If the former is true, incorporating with what we have in (), gcd(k, 4k 1) = gcd (,, n 1) = 1 = gcd(k, n 1) = 1 = gcd(, 4k 1) = gcd (4k 1, n 1) = 1 and is clear to see that (4k 1) = ( n 1) and then k =. If the later holds, then there exists d Z + such that (4k 1) = d( n 1). (3) As (4k 1) and ( n 1) are both odd, this implies d is an odd integer too. Substituting (3) into (1) we have, ( 4k 1)(k) = (k) ( n 1) d= ( n 1). This implies (k) d = and either d = k factors of Q d = k Z+ or d = k Z+ and is an even integer for the only are multiples of, which is a contradiction to d is an odd integer.

Perfect if and only if Triangular 47 Hence( n 1) (4k 1 ) only when d = 1 and hence (4k 1) = ( n 1) and k =. Consequently, (k)( 4k 1) = ( n 1) ( 4k 1) = ( n 1)and k = 4k = n and k = n k = n for some prime n. Thus m = 4k 1 = n 1 = ( n 1), and if an even Triangular number T m is perfect, m = ( t 1) for some prime number t. Alternative proof: (k)(4k 1) = ( n 1) 8k k = ( n 1) 4k k = n ( n 1) 4k k - n ( n 1) = 0 4k k - n ( n 1) = 0 4k n k + ( n 1)k - n ( n 1) = 0 4k(k n ) + ( n 1)(k - n ) = 0 (k n )(4k + n 1) = 0 k = n or 4k = 1 n k = n or k = n = (1 n ) k = n or k = n. k ε (because (1 n ) < 0, n 3 and k = 1 n Z + ). Consequently, m = 4k 1 = 4( n ) 1= ( n 1), where n is prime. Conversely, suppose T m is an even triangular number where m = ( t 1) for some prime number t. Then T m = T t 1 = t 1 i=1 i = (t 1 )( t 1+1) = ( t 1 )( t 1 ) = N and which is perfect. Corollary 14: An even triangular T M is perfect if and only if M is a Mersenne prime. Theorem 15: An even triangular number T 4k 1 is not perfect if k 1 mod 5 or k 0 mod 5. Proof: Consider an even triangular number T 4k 1. Suppose k 1 mod 5 or

48 Tilahun Muche, Mulatu Lemma, George Tessema and Agegnehu Atena k 0 mod 5. Then k 1 mod 5 if 5 (k 4) if and only if k = 4 + 5t for some t Z +. Hence T 4k 1 = T 4( 4+5t) 1 = T 15+0 t = = 0(3+4t)(4+5t) if and only if k 4 mod 5 if and only ( 15+0t)( 0t+16) = 10( 3 + 4t)( 4 + 5t) and 10 T 4k 1. Consequently T 4k 1 is not a perfect number. (Proposition 10). Similarly if perfect. k 0 mod 5 one can show that 10 T 4k 1 which implies it is not Theorem 16 : If a triangular number T m is perfect, then T ( n 1) = T (n+1 ) - 3 T ( ) ( ) = (k 1) 3 k=1. Proof: If a triangular number T m is perfect then m = ( n 1) where n is a prime number (Theorem 13). T n = n i=1 i 3 = ( n(n+1) ). T (n+1 ) n+1 ( ) = i 3 = ( ( i=1 n+1 ) ( (n+1 ) +1) ) = (n+1) ( (n+1 ) +1) 4 = ( n+1 + 1) and (4) T ( ) ( ) = i 3 = ( ( i=1 ) ( ( ) +1) ) = () ( ( ) +1) 4 = (n 3) ( ( ) + 1) 3 T ( ) = 3. (n 3) ( ( ) + 1) = n ( ( ) + 1) (5)

Perfect if and only if Triangular 49 Combining (4) and (5) we have, T (n+1 ) - T ( ) = () ( (n+1 ) + 1) - n ( ( ) + 1) = ( ) (( (n+1 ) + 1) ( ( ) + 1) ) = ( ) (( n+1 +. (n+1 ) + 1) 1 ( +. ( ) + 1)) = ( ) (( n+1 +. (n+3 ) + 1 ) - 1 ( + (n+1 ) + 1)) = ( ) ( n+1 + (n+3 ) + 1 n (n+3 ) ) = ( )( n+1 n 1) = ( )(. n n 1) = ( )( n 1) = n ( n 1) = T ( n 1) (6) ( ) Next we show that, T ( n 1) = (i 1) 3 T ( n 1) = T ( n+1 ) - 3 T ( ) n+1 ( ) i=1. ( ) = (i) 3-3 (i) 3 i=1 i=1 = (1 + 3 + 3 3 + + ( (n+1 ) ) 3) - 3 (1 + 3 + 3 3 + + ( ( ) 3 ) ) = (1 + 3 + 3 3 + + ( (n+1 ) ) 3) - ( 3 + 4 3 + 6 3 + + ( (n+1 ) 3 ) ) = (1 + 3 + 3 3 + 4 3 + 5 3 + 6 3 + 7 3 + 8 3 + + ( n+1 3 ) + ( (n+1 ) 3 1) + ( n+1 3 ) ) - ( 3 + 4 3 + 6 3 + 8 3 + ( n+1 3 ) + ( n+1 3 ) ) = (1 + 3 3 + 5 3 + 7 3 + + ( (n+1 ) 1) 3) = (1 + 3 3 + 5 3 + 7 3 + + (. ( ) 3 1) ) = (1 + 3 3 + 5 3 + 7 3 + + (k 1) 3 ) where k = ( ) ( ) = (k 1) 3 k=1 This implies T n 1 = ( k=1 (k 1) 3. (7) )

50 Tilahun Muche, Mulatu Lemma, George Tessema and Agegnehu Atena From (6) and (7) it follows, T ( n 1) = T (n+1 )- 3 T ( ) ( ) = (k 1) 3 k=1. REFERENCES [1] Vladimir, S., (010), On Perfect and Near-Perfect Numbers, ArXiv.org.math.arXiv: 1011.6160, Mathematics, Number Theory. [] David, M. Burton, Elementary Number Theory, University of New Hampshire, Allyn and Bacon, Inc. Boston, 1980. [3] Stan Wagon, Perfect numbers, Math. Intelligencer 7 (1985), 66 68. [4] G.H. Hardy, and E.M., Wright, An introduction to theory of Numbers. Fourth edition, 1975, Oxford University Press, Ely House, London W. 1. Mersenne Prime Search, http://www.mersenne.org/ Mersenne Primes: History, Theorems and Lists http://primes.utm.edu/mersenne. [5] Muche, T., and Atena A., (106), Investigating Triangular Numbers with greatest integer function Sequences and Double Factorial, APJMR Vol. 4,134-14. [6] Bhabesh D. (013). Some aspect of Perfect Numbers and Generalized Perfect. International Journal of Mathematical Archive-4(6), 013, 118-13. [7] Nihal B., and Recep G., (010). On Perfect Numbers and their Relations. Int. J. Contemp. Math. Sciences, Vol. 5, 010, no. 7, 1337 1346. [8] Azizul H., and Himashree K., (014). Generalized Perfect Numbers Connected with Arithmetic Functions. Math. Sci. Lett. 3, No. 3, 49-53 (014). [9] The on-line encyclopedia of integer sequences. http://www.research.att.com /njas/sequences.

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